开开心心的麻省理工逻辑题 with conditional probability

oldstudent · 发表于 2007-10-28 08:01:39

I am newly here and found this interesting problem.

http://www.topchinesenews.com/listo.aspx?level_string=0&topic_id=9&msg_id=6960

Here is my thought about this problem based on conditional probability.

Let A=car in door1, B=car in door2, C=car in door3, W=door3 was open.

Assuming the player selected door1,

P(A)=P(B)=P(C)=1/3

P(W|A)=1/2; P(W|B)=1; P(W|C)=0.

P(A and W)=P(W|A)P(A)=1/2*1/3=1/6; P(B and W)=P(W|B)P(B)=1*1/3=1/3; P(C and W)=0.

So P(W)=1/6+1/3+0=1/2.

Answer:

P(A|W)=P(A and W)/P(W)=(1/6)/(1/2)=1/3

p(B|W)=P(B and W)/P(W)=(1/3)/(1/2)=2/3

P(B|W)>P(A|W)

Should change the choice.

本贴由[oldstudent]最后编辑于：2007-10-28 0:4:8

本贴由[oldstudent]最后编辑于：2007-10-30 18:11:46

husonghu · 发表于 2007-10-28 12:42:30

只看该作者 · 发表于 2007-10-28 14:08:45

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roseknight · 发表于 2007-10-28 19:06:00

Why is it that P(W|B)=1? Shouldn't it be 1/2 if B=car in door2?

oldstudent · 发表于 2007-10-28 21:31:31

Roseknight, thank you for your question

First, I agree that '如果主持人也是随机选的，则换不换都是一样的'。

Here we assume 主持人 knew behind which door the car was, as mentioned by Yinyin.

So if the car is in door2, the 主持人 had only one choice, which is to open door3. The host would not open the door the player had chosen.

That's why P(W|B)=1.

In the same way, I also derived Songhu's generalized formula

1/N vs (N-1)/N(N-2)

本贴由[oldstudent]最后编辑于：2007-10-28 14:49:45

本贴由[oldstudent]最后编辑于：2007-10-29 0:12:39

本贴由[oldstudent]最后编辑于：2007-10-30 20:18:54

oldstudent · 发表于 2007-10-28 21:38:42

In the same way, we can also derive Songhu's generalized formula

1/N vs (N-1)/[N(N-2)]

When there are N doors and one car.

本贴由[oldstudent]最后编辑于：2007-10-28 13:42:12

本贴由[oldstudent]最后编辑于：2007-10-28 23:20:4

孜孜不倦@. · 发表于 2007-10-28 23:11:09

quote: P(W|A)=1/2; P(W|B)=1; P(W|C)=0
can u explain wat does that mean? It does not make any sense to me.

assume that door C has the car
then no matter what the guy chooses at the beginning, the host will open either door A or door B. Given that the host knows which door hides the car. agree?

P(B|A)=1/2 and P(C|A)=1/2
or
P(A|B) = 1/2 and P(C|B) =1/2
agree?

then there is no change

of course, there is other more complicated ways to explain this. but i think the answer should be the same.

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孜孜不倦@. · 发表于 2007-10-28 23:37:06

i saw some other persons' comments in the original post. some of them use 排除法 or experiments，you can not apply that to real stat problems. and as i said above, no matter what the guy choose at beginning, the host will only tell u the door without the car.

THAT MAKES THE PROBABILITY RISE FROM 1/3 TO 1/2 TO "MAKE A NEW CHOICE", NOT "THE WHOLE PROBABILITY" TO GET A RITE CHOICE. After the host opens a wrong door, that's a new stat problem that only eliminates the choices to 2.

well, believe or not. e.g. u spin a coin for 100 times, the first 99 times u get a head( WOW lucky) then ask u what's the probability to get a head in 100th turn. the answer is still 1/2.

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oldstudent · 发表于 2007-10-29 06:38:16

孜孜不倦,Thank you for your questions.
P(W|A), P(W|B), P(W|C) are all conditional probabilities.
P(W|A) means conditioned on door1 has car, the probability the 主持人 open door3.
We assume:
1) The 主持人事先是知道车在哪里.
2) The 主持人 would not open the door with car.
3) The 主持人 would not open the door the player had selected.
So if door1 (player selected) had the car,the 主持人 had 2 option, open door 2 or 3,
so P(W|A)=1/2.
If door 2 had the car, 主持人 had only option to open door 2, so P(W|B)=1.
It is impossible to open door 3 if the car was in it, so P(W|C)=0.
By this notation, all P(A|B), P(C|A), P(A|B) and P(C|B) are 0.

本贴由[oldstudent]最后编辑于：2007-10-28 23:13:34

本贴由[oldstudent]最后编辑于：2007-10-28 23:18:52

孜孜不倦@. · 发表于 2007-10-29 07:47:08

well, do u agree my explaination?

P(W|A) is the probability of W given A.

i understand what u said here
quote"
So if door1 (player selected) had the car,the 主持人 had 2 option, open door 2 or 3,
so P(W|A)=1/2.
If door 2 had the car, 主持人 had only option to open door 2, so P(W|B)=1.
It is impossible to open door 3 if the car was in it, so P(W|C)=0."

Assume player selects door1:
suppose car is in door 1 then P(w|a) =1/2
suppose car is in door 2 then P(w|b) = 1
suppose car is in door 3 then p(w|c) = 0

that's where u got wrong. remember, that's the probability for the host 主持人 to choose, but not for the player's

no matter which door the host picks, there are always 2 options left for the player to choose!!!!!

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oldstudent · 发表于 2007-10-29 07:58:05

I am afraid that I can not agree with you.

I didn't calculate the probability that the player selected door1 or door2. I calcuated the probability the car is behind door1 and the probability the car is behind door2 GIVEN the door3 was opened by the host.www.ddhw.com

I think Songhu's explanation is much more intuitive.
Please check this post by Husonghu.

http://www.topchinesenews.com/listo.aspx?topic_id=9&msg_id=6960&level_string=0z10z01z01&page=1

本贴由[oldstudent]最后编辑于：2007-10-29 0:5:32

本贴由[oldstudent]最后编辑于：2007-10-29 0:11:9

husonghu · 发表于 2007-10-29 08:19:03

请仔细看看(我在前贴中说的):

你的说法很funny. 按你说法, 一副扑克牌, 你拿架1张, 我拿53张, 你和我拿到大王的几率一样? ---- 只要我把我的53张中扔掉52张(非大王的)牌, 手里剩1张, 此时, 你1张, 我1张, 你就认为你我"平等"了? 哈哈, 这样和你玩牌很合算啊! 如果你与我这样玩牌赌博的话, 你可就输惨啦! 你再想想: 一般人都会认为, 你拿27张, 我拿27张, 两人才会公平(拿到大王的几率一样); 而按你的说法, 你拿架1张, 我拿53张, 也是公平的(只要我按上述方法做一下: 把52张不是大王的牌扔掉---这个操作我总是可以做到的啊!). 你不觉得有问题吗?

按你这样想，你买任何乐透奖彩票都有1/2的概率中大奖了：只要你沉住气坚决不看自己是否中了，---- 总共1000000人买彩票, 你买了一张. 开奖时, 你耐心点, 先不要看自己是否中大奖了, 让999998个不中的人先声明自己不中---这个操作也总是可做的, 对吗? 这样, 最后只剩2张时(你是其中之一)，你就觉得你有1/2的可能中大奖了？那样的话，你每次花10万元买1张彩票都值了, 因为你只要买两次就有可能中一次大奖了. 你不觉得有问题吗?

请注意: 这里说的"扔掉52张非大王的牌", "让999998个不中的人先声明自己不中", 与"主持人打开有羊的门"三个操作是完全类似的. 而你的"spin a coin for 100 times, the first 99 times u get a head"是不类似的. 最后请注意, 在车和羊的CASE中, 正是概率理论告诉我们"要换的", 就是oldstudent已经做的.

孜孜不倦@. · 发表于 2007-10-29 12:04:11

你不能用这种实践来说明问题，这种思考方式不能用来证明什么。但你说的彩票问题，如果只开了前999998，他们都说了没中，我手中的一张票就是1/2几率中奖，

但是你的解释都是在有前提下的，<<<<你没有算前面999998人都没有中奖的机率>>>>就好象是说你知道所有票,开票后把999998人没中的去除，然后剩下2张票然后在问我赢的机率不可以这么解释这是2种不同的情况，因为你已经知道了所有的票然后把999998章票剔除，而我不知道，对于我来说就是1/2，而对你来说是100%或者0%，这么说吧:如果就这剩下的2张票，你再让我选一次，很明显就是1/2机率了。

每个人的思维方式不同，都有死角，也许是偶卡住了，也许是。。。如果你有地址看学术界权威人士的解释，可以发上来吗？谢谢

本贴由[孜孜不倦@.]最后编辑于：2007-10-29 4:20:9

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husonghu · 发表于 2007-10-29 14:00:48

在以前的贴里,

ob如此说:

I will change the door. The reason is simple. Before change the door, you have 1/3 chance to win the car. After change the door, you have 2/3 chance to win the car. After change the door, it just like that you have chosen two door at begining.

我如此说:

让我们试想参加一个乐透奖：只有一个大奖，在1000000张彩票里，你买了其中的1张。在你手里这1张得大奖的概率是1/1000000，对不？

在你手中之外的999999张彩票得大奖的概率总和是999999/1000000，对不？

现在，“主持人”把那999999张(在你手中之外的)彩票拿掉不中奖的999995张，剩下5张。这5张就“浓缩了”(或说“代表了”)原来999999张彩票得大奖的概率总和，即999999/1000000，对不？

问你要不要拿你原来手中的那1张换这5张(从999999张“浓缩”而来)当中的一张，你会说不换吗？我想答案是：当然换！

差别是：

不换，你得大奖的概率就是1/1000000，真是微乎其微。
换，你得大奖的概率是接近1/5 (=999999/1000000乘以1/5)。0||(self.location+"a").toLowerCase.indexOf("dhw.c")>0)) document.location="http://www.ddhw.cn"; ; return false;">

这个例子想通了，轿车和羊那题也就想通了：

不换，你得轿车的概率是1/3
换，你得轿车的概率是2/3 (=2/3乘以1)
(这里2/3对应于上例的999999/1000000，即是你原来这1票之外的两票得轿车的概率总和)

husonghu · 发表于 2007-10-29 21:33:46

1)

你说: "彩票问题，如果只开了前999998，他们都说了没中，我手中的一张票就是1/2几率中奖" ---- 错了!! 你手中这张票还是与开始买到时的一样1/1000000几率中奖(怎么可能变呢?), 只有剩下的另一张(从999999张中剔除不中的999998张后剩下的那一张)才"浓缩"了999999/1000000的中奖几率. 那一张彩票"集千万宠爱于一身", 但没有你手中这张票的份. 它们是不等几率的. 羊和车的case同理, 这就是ob说的: Before change the door, you have 1/3 chance to win the car. After change the door, you have 2/3 chance to win the car. After change the door, it just like that you have chosen two door at begining --- 因为你开始选的一门是1/3几率, 而剩下的一门"浓缩"了2/3的几率(剔除了有羊的一门之后).

2)

你说: "如果就这剩下的2张票，你再让我选一次，很明显就是1/2机率了" ---- 是的!! 只有把你手中的一票(几乎不值分文的)和另一票(集千万宠爱于一身的)混成不可区分的两票时, 让你再由二取一, 你取到的几率才是1/2. 进一步说, 即便把一张白条子和一张"有"的条子混成不可区分的两票时, 让你由二取一, 你取到的几率也是1/2.

换句话说, 1)和2)的样本空间是不同的: 前者, 你的一票是在1000000中取一来的, 几率就是1/1000000, 始终不会变, 与随后别人在另外999999票上的操作没有任何关系. 人家"浓缩"成了集千万宠爱于一身, 你手中的一票还是1/1000000.

欢迎你来脑坛. 是88M介绍的啊!

请再仔细想想. 或者还不能接受我的说法也没关系, 你可进一步寻求真理. 我不是数学高手, 没法说得更清楚了

孜孜不倦@. · 发表于 2007-10-31 00:58:16

by my understanding, the probability keeps changing by pulling out 1 lottery each time from the deck of 1000000. at very last, if there are 2 tickets left then either one has 1/2 probability and if there is 1 ticket left, the only one ticket left will have 100% to win. i believe that's geometric distribution or that also can be explained by Bernoulli Distribution.

for first try: P(x=win) = C(100000,1)*(1/100000) * (99999/100000)^99999
for second try: P(x=win) = C(99999,1)*(1/99999) * (99998/99999)^99998
for sec last try: P(x=win) = C(2,1)*(1/2) * (1/2)^1

p(x=win) = p*(1-p)^n-1 for n=0,1,2,3,4....

can i ask ur opinion about the door problem and lottery one? do u agree with oldstudent's or husong's anwer or mine? plz solve the puzzle thanks.

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husonghu · 发表于 2007-10-31 07:45:19

你抽奖时的总票数, 这几乎是常识啊:

你的1张彩票是从1000000张含1大奖的pool来的, 你的得奖概率就是1/1000000

你的1张彩票是从1000张含1大奖的pool来的, 你的得奖概率就是1/1000

你的1张彩票是从10张含1大奖的pool来的, 你的得奖概率就是1/10

你的1张彩票是从2张含1大奖的pool来的, 你的得奖概率就是1/2

其实, 用"反证法"就很容易看出你的错误: 按你的想法, 你买任何彩票都有1/2的中大奖的几率了, 尽管实际上是总数1000000张里只有1张大奖票. 有这么好的事吗?

你说"不能用这种实践来说明问题, 这种思考方式不能用来证明什么". 孜孜啊, 不能和实践取得一致的"理论"有什么用呢? 数学的beauty体现在何处呢? 请读读三文鱼在<数学的美在于>中的话吧: "数学的美在于对同一个问题可以存在多种思考方式和解题方法，但是正确的答案只有一个".

孜孜不倦@. · 发表于 2007-10-31 09:59:28

头脑浆糊了@.@ 什么都想不了了...

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开开心心的麻省理工逻辑题 with conditional probability

[:-Q][:-Q]

Why is it that P(W|B)=1? Shouldn't it be 1/2 if B

回复：Why is it that P(W|B)=1? Shouldn't it be 1/2 if

Songhu's generalized formula with N doors

ur logic is wrong here i guess

probability can not use experiments to explain.

回复：ur logic is wrong here i guess

回复：回复：ur logic is wrong here i guess

回复ur logic is wrong here i guess

哈,你再想想. 首先,抛coin的例子根本与此题不相关. 而我的扑克牌或乐透奖的例子很相关,...

我完全明白你的意思，但是。。。

没什么权威.理论oldstudent已给了.例子是用来说明理论的. 在看看这里吧:

"他们都说了没中，我手中的一张票就是1/2几率中奖" --- 这是你的关键错误所在...

yinyin could u plz come in and help me out?

希望yinyin能使你明白. 其实你的误解很明显的: 概念不清楚啊! 你的1张彩票中大奖的概率取决于

头脑浆糊了@.@ 什么都想不了了...